$$\lim_{x \to \frac{\pi}{2}} \frac{\cot x - \cos x}{(\pi - 2x)^{3}}$$ equals

We have to evaluate the limit

$$\lim_{x \to \frac{\pi}{2}} \dfrac{\cot x - \cos x}{(\pi - 2x)^{3}}.$$

The point toward which $$x$$ approaches is $$\dfrac{\pi}{2}$$, so it is convenient to translate the variable. We write

$$x \;=\;\dfrac{\pi}{2} + h \qquad\text{with}\qquad h \rightarrow 0.$$

First let us rewrite the denominator in terms of $$h$$:

$$\pi - 2x \;=\;\pi \;-\;2\left(\dfrac{\pi}{2}+h\right)=\pi-\pi-2h=-2h,$$

so

$$(\pi - 2x)^{3}=(-2h)^{3}=-8h^{3}.$$

Next we expand the trigonometric functions near $$h=0$$. We shall use the standard Taylor series (about $$0$$):

$$\sin t = t - \dfrac{t^{3}}{6} + \dfrac{t^{5}}{120} - \dots,$$

$$\cos t = 1 - \dfrac{t^{2}}{2} + \dfrac{t^{4}}{24} - \dots.$$

Because $$x=\dfrac{\pi}{2}+h,$$ we have

$$\sin x = \sin\!\left(\dfrac{\pi}{2}+h\right)=\cos h = 1 - \dfrac{h^{2}}{2} + \dfrac{h^{4}}{24} - \dots,$$

$$\cos x = \cos\!\left(\dfrac{\pi}{2}+h\right)= -\sin h = -h + \dfrac{h^{3}}{6} - \dfrac{h^{5}}{120} + \dots.$$

Since $$\cot x = \dfrac{\cos x}{\sin x},$$ we now form that quotient. Let

$$N=-h + \dfrac{h^{3}}{6} - \dfrac{h^{5}}{120} + \dots \qquad\text{and}\qquad D=1 - \dfrac{h^{2}}{2} + \dfrac{h^{4}}{24} - \dots.$$

We need $$\dfrac{N}{D}$$ up to the term containing $$h^{3}$$. Because

$$\dfrac{1}{1-z}=1+z+z^{2}+\dots,$$

we replace $$z$$ by $$\dfrac{h^{2}}{2}-\dfrac{h^{4}}{24}+\dots$$ and obtain

$$\dfrac{1}{D}=1+\dfrac{h^{2}}{2}+O(h^{4}).$$

Multiplying this by $$N$$ gives

$$$ \begin{aligned} \cot x&=\left(-h+\dfrac{h^{3}}{6}+O(h^{5})\right) \left(1+\dfrac{h^{2}}{2}+O(h^{4})\right)\\ &= -h\Bigl(1+\dfrac{h^{2}}{2}\Bigr)+\dfrac{h^{3}}{6}+O(h^{5})\\ &= -h-\dfrac{h^{3}}{2}+\dfrac{h^{3}}{6}+O(h^{5})\\ &= -h-\dfrac{h^{3}}{3}+O(h^{5}). \end{aligned} $$$

Hence we have

$$\cot x = -h - \dfrac{h^{3}}{3} + O(h^{5})$$

and, from above,

$$\cos x = -h + \dfrac{h^{3}}{6} + O(h^{5}).$$

Now we form the numerator of the original fraction:

$$$ \begin{aligned} \cot x - \cos x &=\left(-h - \dfrac{h^{3}}{3}+O(h^{5})\right) -\left(-h + \dfrac{h^{3}}{6}+O(h^{5})\right)\\ &= -h + h -\dfrac{h^{3}}{3}-\dfrac{h^{3}}{6}+O(h^{5})\\ &= -\dfrac{h^{3}}{2}+O(h^{5}). \end{aligned} $$$

So the entire expression becomes

$$$ \frac{\cot x-\cos x}{(\pi-2x)^{3}} =\frac{-\dfrac{h^{3}}{2}+O(h^{5})}{-8h^{3}} =\frac{1}{2}\cdot\frac{1}{8}+O(h^{2}) =\frac{1}{16}+O(h^{2}). $$$

As $$h \rightarrow 0,$$ the error term $$O(h^{2}) \rightarrow 0,$$ and we obtain the required limit:

$$\lim_{x \to \frac{\pi}{2}} \dfrac{\cot x - \cos x}{(\pi - 2x)^{3}} = \frac{1}{16}.$$

Hence, the correct answer is Option B.